Homework 11/Solutions. (Section 6.8 Exercise 3). Which pairs of the following vector spaces are isomorphic?

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1 MTH 9-4 Linear Algebra I F Section Exercises 6.8,4,5 7.,b 7.,, Homework /Solutions (Section 6.8 Exercise ). Which pairs of the following vector spaces are isomorphic? R 7, R, M(, ), M(, 4), M(4, ), P 6, P 8, P, P Since dim R n = n, dim M(m, n) = mn and dim P n = n+ we obtain the following chart: V R 7 R M(, ) M(, 4) M(4, ) P 6 P 8 P P dim V By Theorem 6. two finite dimensional vector spaces are isomorphic if and only if they have the same dimension. Also a finite dimensional vectors space is not isomorphic to an infinite dimensional vector space. So we obtain the following (unordered) pairs of isomorphic vector spaces dim V = 7 : {R 7, P 6 } dim V = 9 : {R 9, M(, )} dim V = : {R, M(, 4)}, {R, P }, {P, M(, 4)}, {R, M(4, )}, {P, M(4, )}, {M(, 4), M(4, )} (Section 6.8 Exercise 4). (a). For any vector space V, show that id V : V V is an isomorphism. (b) Suppose T : V V is an isomorphism from the vector space V to the vector space V. Prove that T is invertible and that T is an isomorphism from V to V. (c) Suppose T : V V and T : V V are isomorphisms. Prove that T T : V V is an isomorphism. (a) By Lemma A.5. in the appendix of the notes, id V id V = id V. So id V is an inverse of id V. Also by Section 6. Exercise 6, id V is linear and so id V is an isomorphism. (b) This is obvious with the definition of an isomorphism in the notes. (But observe that according to Theorem 6.8 in the notes, the definition of an isomorphism in the notes is equivalent to the definition in the book). (c) By (b) T and T are invertible and so by A.5.6 in the appendix of the notes, T T is invertible. By Theorem 6.7, T T is invertible and so by Theorem 6.8, T T is an isomorphism.

2 (Section 6.8 Exercise 5). (a) Show that any vector space V is isomorphic to itself. (b) Show that if a vector space V is isomorphic to a vector space V, then V is isomorphic to V. (c) Show that if the vector space V is isomorphic to the vector space V and V is isomorphic to the vector space V, then V is isomorphic to V. Recall call first that by definition a vector space V is isomorphic a vector space W if and only if there exists an isomorphism T : V W. (a) By Section 6.8 Exercise 4a, id V : V V is an isomorphism. So V is isomorphic to V. (b) Suppose that the vector space V is isomorphic to the vector space V. Then there there exists an isomorphism T : V V. By Section 6.8 Exercise 4b, T : V V is an isomorphism and so V is isomorphic to V. (c) Suppose the vector space V is isomorphic to the vector space V and V is isomorphic to the vector space V. Then there exist an isomorphism T : V V and an isomorphism T : V V. By Section 6.8 Exercise 4c, T T : V V is an isomorphism and so V is isomorphic to V. 7 (Section 7. Exercise b). Use Theorem 7.4 to compute det 9 5 by keeping 4 8 track of the changes that occur as you apply row operations to put the matrix in reduced row-echelon form R4 + R R4 R R R R R R 5 5 R4 R4 R R R R4 66 So the determinant is R + R R R R R R4 5R R4 R R4 R4 R + 7 R R 66 7 (Section 7. Exercise ). Suppose A is a square matrix. Use induction to prove for any integer n, that det A n = (det A) n.

3 Recall first that A n is inductively defined by A = I, and A n+ = A n A By definition of a regular determinant function det I =. Thus det A = det I = = (det A) and so det A n = (det A) n holds for n =. Suppose now that ( ) det A n = (det A) n for some non-negative integer n. Then det A n+ = det A n A definition of A n+ = det A n det A Theorem 7.7 = (det A) n det A by the induction assumption ( ) = (det A) n+ Property of R So det A n+ = (det A) n+. Thus by the principal of induction, det A n = (det A) n holds for all non-negative integers n. (Section 7. Exercise ). Prove that if the square matrix A is invertible, then det A = (det A). Since A is invertible, AA = I. Thus by Theorem 7.7, det(a) det(a ) = det(aa ) = det I. By definition of a regular determinant function det I = and so det(a) det(a ) =. Thus det A = (det A). (Section 7. Exercise ). Prove that if A and P are n n- matrices and P is invertible, then det(p AP ) = det A Using Theorem 7.7 twice we compute ( ) det(p AP ) = det P (AP ) = det P det(ap ) = det P det A det P = det P det P det A By Section 7. Exercise, det(p ) = (det P ) and so det P det P =. Thus det(p AP ) = det A. A. Fill in all the? in the proof of the following Theorem: Theorem A. Let A be a m n matrix and B its reduced row echelon form. Let x f,..., x ft be the free variables of B and let s be number of non-zero rows of B. Let (e,..., e n ) be the standard basis for R n and let b k be row k of B. Then (b,..., b s ) is a basis for RowA and (b,..., b s, e f,..., e ft ) is basis for R n.

4 Proof. Put D = (b,..., b s, e f,..., e ft ) Note that (b,..., b s ) is the list of non-zero rows of B. By Theorem N.7.5 (b,..., b s ) is a basis for RowA. So we just need to show that D is a basis for R n. Note that s is the number lead variables and so n = s + t. Thus D is a list of length n in the n -dimensional vector space R n. So by Theorem N.5.5 ( ) D is basis of R n if and only if D is linearly independent. To show that D is linearly independent, let r,..., r s, u,..., u t R such that ( ) r b r s b s + u e f u t e ft = Let k s and let b klk be the leading in b k. Then b klk is the only non-zero entry in Column l k of B and so the l k entry of b j is for all j s with j k. Since x lk is a leading variable, l k f j for all j t and so also the l k entry of e fj is. Thus the l k entry of the linear combination on the left side of the equation ( ) is r k. Hence r k = for all k s. Thus ( ) implies u e f u t e ft = Since (e,..., e n ) is a basis and so linearly independent this gives u j = for all j t. Thus D is linearly independent, and so by ( ) D is a basis for R n. ( ) B. Let V = span (,,,, ), (,,,, ), (,,,, ). Find a basis for V and extend it to a basis of R 5. Hint: Use Theorem A to find both bases simultaneously. We use the Gauss-Jordan Algorithm to compute the reduced row echelon form of the matrix A formed by the the above list of vectors as rows. R R R R R R R R R R R Thus by Theorem I, 4

5 is a basis for ColA = V and since x, x 4, x 5 are the free variables is a basis for R 5. 5